A finite abelian group is the internal direct product of its primary components

Given: The objects and hypotheses in the statement.

Let $G$ be finite abelian and write $\mathrm{\mid }G\mathrm{\mid }=p^{a}m$ with $p\nmid m$. Then $G(p)$ is a subgroup of order $p^a$. It is the unique subgroup of $G$ having that order. In particular, if $p\nmid \mathrm{\mid }G\mathrm{\mid }$, then $G(p)=\{e\}$. (A p-primary component has the full p-power order and is the unique subgroup of that order).

Let $N_0,\dots ,N_{r-1}⊴G$. The following are equivalent: the $N_i$ form an internal direct product of $G$; every $g\in G$ has a unique expression $g=n_0\cdots n_{r-1}$ with $n_i\in N_i$; and the multiplication map $\mu :{\prod }_{i<r}{N}_i\to G$ is an isomorphism. These statements include the empty family and the one-factor case. (Internal direct products are external direct products, equivalently every element has a unique factorisation).

Powers are the natural powers of def-group-power and finite products those of def-monoid-finite-product, both taken in the commutative monoid $(\mathbb{Z},\cdot ,1)$ of lem-units-of-z. Call $p:r\to \mathbb{Z}$ an injective list of primes when every $p_i$ is prime (def-prime) and $p_i=p_j$ forces $i=j$ (def-injection-surjection-bijection). Let $n\in \mathbb{Z}$ with $n\ge 1$ and let $p:r\to \mathbb{Z}$ be an injective list of primes such that every prime divisor of $n$ equals $p_i$ for some $i<r$. Then, with $v_q$ as in def-p-adic-valuation: 1. ${\displaystyle n=\prod _{i<r}{p}_i^{v_{p_i}(n)}}$; 2. $v_q(n)=0$ for every prime $q$ that is not among $p_0,\dots ,p_{r-1}$; 3. the exponents are determined by $n$: if $e:r\to \mathbb{N}$ and $n={\prod }_{i<r}{p}_i^{e_i}$, then $e_j=v_{p_j}(n)$ for every $j<r$. Clause 3 needs only injectivity of the list, not the covering hypothesis. (For $n\ge 1$ and any injective list $p:r\to \mathbb{Z}$ of primes containing every prime divisor of $n$, one has $n={\prod }_{i<r}{p}_i^{v_{p_i}(n)}$; the exponents are determined by $n$, and $v_q(n)=0$ for every prime $q$ outside the list).

Let $a,b\in \mathbb{Z}$, not both $0$, and put $I:=\{ax+by:x,y\in \mathbb{Z}\}\mathrm{.}$ Then $I$ contains a positive element, and its least positive element is $\gcd (a,b)$ (def-common-divisor-and-gcd). In particular there are integers $x_0,y_0$ with $a{x}_0+b{y}_0=\gcd (a,b),$ so the equation $ax+by=\gcd (a,b)$ is solvable in $\mathbb{Z}$. (Bézout's identity: for integers $a,b$ not both zero, $\gcd (a,b)$ is the least positive element of $\{ax+by:x,y\in \mathbb{Z}\}$; in particular $ax+by=\gcd (a,b)$ has an integer solution).

If $G$ and $H$ are finite groups, then their external direct product is finite and has order $\mathrm{\mid }G\times H\mathrm{\mid }=\mathrm{\mid }G\mathrm{\mid }\mathrm{\mid }H\mathrm{\mid }$. (For finite groups $G$ and $H$, $\mathrm{\mid }G\times H\mathrm{\mid }=\mathrm{\mid }G\mathrm{\mid }\mathrm{\mid }H\mathrm{\mid }$).

Let $G$ be a finite group and $H\le G$. Then $\mathrm{\mid }G\mathrm{\mid }=[G:H]\mathrm{\mid }H\mathrm{\mid }\mathrm{.}$ Consequently, under the canonical embedding $\iota :\mathbb{N}\to \mathbb{Z}$, $\mathrm{\mid }H\mathrm{\mid }$ divides $\mathrm{\mid }G\mathrm{\mid }$. (Lagrange's theorem: $\mathrm{\mid }G\mathrm{\mid }=[G:H]\mathrm{\mid }H\mathrm{\mid }$ for every subgroup $H$ of a finite group $G$).